By Lawrence Ernest Bridger
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Extra resources for Rings with a polynomial identity
Example text
J/Q. i s the Jacobson radical of F[x ]/Q. A Ct F[x ]/Q ^^x) • Hence A Ct . I. algebra. I. algebras. I. 2 assures us that F[x ]/J satisfies this a F L x J „ S„ (x ) = 0 in :;— . implies that S« (x ) e J/Q". and since S„ (x ) i s then quasi-regular 2m a A 2n a ,• n modulo Q, i t must be nilpotent modulo Q. e. e. we have 1 j X for some integer and thus, A n > 0 satisfies . same as saying that S„ (x ) = 0 2m a n ^2n^ a^ X n E ^A 24. ) 0, i s an infinite integral domain. (2) aA = 0 for a e 0, implies that (3) a(ab) = (aa)b•= a(ab) a =0 for a l l a e Q, a,b e A With this more general result, we now note that we do not even need ft to be infinite.
I. _ 1 . k be the s m a l l e s t p o s i t i v e i n t e g e r such t h a t t h e r e e x i s t s 39. a set {6^} of elements i n ft , not a l l zero, with the property that a E 3 TT a ... a = '"k ^2 TT^ IT where TT i s a permutation of 1,2,... , arbitrary. E a. x. x. x. = 0 i 1 2 d satisfies the polynomial identity Since we see that 1 k exists and k < d . Z 1 a K~l k . . a 1 TT TT K. a TT_ 1 S 1 TT a TT. TT =k » IT''" for which . This can be rewritten h 1 P TT 1 1 a TT a 1 IT, , TU \ \ - l - 0 k-1 I^K^_^ i s a non-zero two-sided ideal of A A 0 ^ where we are now summing over a l l those permutations TT^Ck) .
This gives us, be a prime P . I . r i n g . Then any n i l s u b r i n g i snilpotent. 1) which s t a t e s t h a t f o r P . I . r i n g s ' n i l i m p l i e s l o c a l l y n i l p o t e n t . I. j ring A finitely If B i s n o t n i l p o t e n t we s h a l l reach a c o n t r a d i c t i o n . generated i m p l i e s that lemma choose, an i d e a l ). prime P . I . r i n g . A i s finitely n generated. By Zorn's maximal r e t h e e x c l u s i o n o f i s a prime i d e a l by a s t a n d a r d us t h a t i t i s n i l p o t e n t .
Rings with a polynomial identity by Lawrence Ernest Bridger
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